英文:
Usage and meaning of &^ and &^= operators in Go
问题
我已经寻找了大约一周，但似乎找不到关于Go语言中这两个运算符&^和&^=的合理解释以及它们的使用方法。是否有人可以友好地给我解释一下呢？
英文:
I've been looking around for about a week now and can't seem to find a decent explanation for these two operators, &^ and &^=, in the Go language and how they are used. Would anybody be as kind as to enlighten me?
答案1
得分: 11
这在我们查看所有位运算符时更容易理解：
& 按位与
| 按位或
^ 按位异或
&^ 按位清除（与非）
- 按位与（
&）：当两个操作数的位都为1时，结果为1，否则结果为0。 - 按位或（
|）：当至少一个操作数的位为1时，结果为1，否则如果两个操作数的位都为0，则结果为0。 - 按位异或（
^）：当且仅当一个操作数的位为1时，结果为1，否则结果为0。这三个运算符（&，|，^）无论操作数的顺序如何，都会产生相同的结果。 - 按位与非（
&^）：当第一个操作数的位为1且第二个操作数的位为0时，结果为1，否则结果为0。注意，操作数的顺序会影响结果。要使结果为1，第一个操作数的位必须为1，第二个操作数的位必须为0。
以下是演示位运算符行为的代码，也可以在Go Playground上找到：
package main
import "fmt"
func main() {
fmt.Println(`AND`)
fmt.Printf("%b & %b 的结果是 %03b\n", 4, 5, 4&5)
fmt.Printf("%b & %b 的结果是 %03b\n", 5, 4, 5&4)
fmt.Println(`OR`)
fmt.Printf("%b | %b 的结果是 %03b\n", 4, 5, 4|5)
fmt.Printf("%b | %b 的结果是 %03b\n", 5, 4, 5|4)
fmt.Println(`XOR`)
fmt.Printf("%b ^ %b 的结果是 %03b\n", 4, 5, 4^5)
fmt.Printf("%b ^ %b 的结果是 %03b\n", 5, 4, 5^4)
fmt.Println(`AND NOT`)
fmt.Printf("%b &^ %b 的结果是 %03b\n", 7, 5, 7&^5)
fmt.Printf("%b &^ %b 的结果是 %03b\n", 5, 7, 5&^7)
}
运行上述代码生成的输出为：
AND
100 & 101 的结果是 100
101 & 100 的结果是 100
OR
100 | 101 的结果是 101
101 | 100 的结果是 101
XOR
100 ^ 101 的结果是 001
101 ^ 100 的结果是 001
AND NOT
111 &^ 101 的结果是 010
101 &^ 111 的结果是 000
最后，&^= 是一个简写的赋值运算符。例如，x = x &^ y 可以替换为 x &^= y。
英文:
This is easier to understand when we take a look at all the bitwise operators:
& bitwise AND
| bitwise OR
^ bitwise XOR
&^ bit clear (AND NOT)
- Bitwise AND (
&): Result is 1 when both operand bits are 1, else the result is 0. - Bitwise OR (
|): Result is 1 when at least one operand bit is 1, else 0 if both operand bits are 0. - Bitwise XOR (
^): Result is 1 when one, and only one operand bit is 1, else the result is 0. These three operators (&, |, ^) produce the same result irrespective of the order of operand bits. - Bitwise AND NOT (
&^): Result is 1 when the first operand bit is 1, and the second operand bit is 0; else the result is 0. Note that the order of the operand bit affects the result. For the result to be 1, the first operand bit must be 1 and the second must be 0.
Here's code, also on the Go Playground, that demonstrates the behavior of bitwise operators:
package main
import "fmt"
func main() {
fmt.Println(`AND`)
fmt.Printf("%b & %b results in %03b\n", 4, 5, 4&5)
fmt.Printf("%b & %b results in %03b\n", 5, 4, 5&4)
fmt.Println(`OR`)
fmt.Printf("%b | %b results in %03b\n", 4, 5, 4|5)
fmt.Printf("%b | %b results in %03b\n", 5, 4, 5|4)
fmt.Println(`XOR`)
fmt.Printf("%b ^ %b results in %03b\n", 4, 5, 4^5)
fmt.Printf("%b ^ %b results in %03b\n", 5, 4, 5^4)
fmt.Println(`AND NOT`)
fmt.Printf("%b &^ %b results in %03b\n", 7, 5, 7&^5)
fmt.Printf("%b &^ %b results in %03b\n", 5, 7, 5&^7)
}
The output generated by running the above code is:
AND
100 & 101 results in 100
101 & 100 results in 100
OR
100 | 101 results in 101
101 | 100 results in 101
XOR
100 ^ 101 results in 001
101 ^ 100 results in 001
AND NOT
111 &^ 101 results in 010
101 &^ 111 results in 000
And finally, &^= is a shorthand assignment operator. For example, x = x &^ y can be replaced by x &^= y
答案2
得分: 1
根据规范，它们是位清除运算符：
&^ 位清除（AND NOT） 整数
你可以将它们用作位标志值的一部分。使用or来打开一个位，使用and not来关闭它。
英文:
The spec says that they are the bit clear operators:
&^ bit clear (AND NOT) integers
You would use them as part of a bit flag value. You'd use or to turn on a bit, and and not to turn it off.
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